Ricci Scalar as Curvature

Question

So I understand that the Ricci scalar represents the curvature of the space. Since any manifold can be considered locally flat, is Ricci scalar always zero locally for any manifold? On one hand it being defined as curvature makes sense it should be. But mathematically it involves the second order derivative of metric tensor which should be zero locally unless the space is globally flat. So I am kinds confused.

Answer 1

The Ricci scalar is a scalar invariant, so it has the same value in all coordinate systems. That means we cannot make a non-zero Ricci scalar zero just by choosing coordinates in which the manifold looks locally flat. Curved spacetimes may have a zero Ricci scalar, for example it is zero for any vacuum solution like the Schwarzschild metric, but if a spacetime has a non-zero Ricci scalar then it is non-zero in all coordinates.

The confusion arises because the phrase locally flat tends to be used rather carelessly. When we say a manifold is locally flat this means that at any point we can choose coordinates (the normal coordinates) in which the Christoffel symbols are zero. However this does not make the manifold flat. Flatness means the Riemann tensor is zero, and switching to normal coordinates does not make the Riemann tensor zero.

Answer 2

The locally flatness means that we can choose a coordinate system at $P$ such that $g_{\alpha\beta}(\mathcal{P}) = \eta_{\alpha\beta} = {\rm diag}(-1,1,1,1)$ and $g_{\alpha\beta,\gamma}(\mathcal{P}) = 0$. In this way, the Christoffel symbols vanishes at $\mathcal{P}$ and so there is no gravitational acceleration. However, we can not make $g_{\alpha\beta,\gamma\delta}(\mathcal{P})$ vanishes generally. Actually, Riemann curvature tensor is a true tensor. If it vanishes in one coordinate system, it will vanish in all coordinate system. In physics, we can eliminate gravitational acceleration, but we can not eliminate the tidal force generated by the inhomogeneity of gravitation field. And this is the difference between gravitation and inertial force.

Answer 3

1. Locally flat appears here to be a layman's characterization that a tangent space $T_pM$ locally approximate a manifold $M$ at a point $p$. This does not imply that the Riemann curvature tensor vanishes at $p$.

2. Concerning the notion of locally flat, there appears to be various mathematical definitions, cf. e.g. Wikipedia and this Math.SE post.

3. Nevertheless, flat in a neighborhood $\Omega$ means that the Riemann curvature tensor vanishes in $\Omega$.

4. Example: The (idealized) surface of the Earth may look flat to an earthling, but it still has a radius of curvature given by $R_E\approx 6\times 10^6 m$, so that that the 2D Ricci scalar (which is twice the Gaussian curvature) is $2/R_E^2$. Indeed small, but technically speaking not flat :)